Equivalence of Definitions of Locally Path-Connected Space/Definition 1 implies Definition 3
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let each point $x$ of $T$ have a local basis $\DD_x$ consisting entirely of path-connected sets in $T$.
Then
- $T$ has a basis consisting of path-connected sets in $T$.
Proof
By definition of local basis, each of these path-connected sets in $\DD_x$ is open in $T$.
Consider the set $\ds \DD = \bigcup_{x \mathop \in S} \DD_x$.
From Union of Local Bases is Basis, $\DD$ is a basis for the topology $\tau$.
Since each $\DD_x$ consists entirely of path-connected sets, $\DD$ also consists entirely of path-connected sets.
$\blacksquare$